# 6 Distribusi Probabilitas Diskrit

Chapter 5

Discrete Probability Distributions

Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-1

Business Statistics: A Decision-Making Approach

8th Edition

Chapter Goals

After completing this chapter, you should be able to:

Calculate and interpret the expected value of a discrete

probability distribution

Apply the binomial distribution to business problems

Compute probabilities for the Poisson and

hypergeometric distributions

Recognize when to apply discrete probability

distributions to decision making situations


Introduction to Probability Distributions

Probability is the way decision makers express their

uncertainty about outcomes and events.

Random Variables A variable that takes on different numerical values based on chance.

Represents a possible numerical value from a random event.

Can vary from trial to trial.

Contoh : Inspektor memeriksa 3 TV plasma dan menentukan ‘diterima’ atau

‘tidak diterima’.

Outcome/hasil dari eksperimen yang mungkin adalah :

x = {0, 1, 2, 3} random variable


Introduction to Probability Distributions


Random

Variables

Discrete

Random Variable

Continuous

Random VariableCh. 5 Ch. 6

Discrete Random Variable

A discrete random variable is a variable that can assume only

a countable number of values

Many possible outcomes:

number of complaints per day

number of TV’s in a household

number of rings before the phone is answered

Only two possible outcomes:

gender: male or female

defective: yes or no

spreads peanut butter first vs. spreads jelly first


Contoh :

Kuis Statistik Industri 1 berupa soal pilihan ganda. Apabila x

merepresentasikan jumlah pertanyaan yang dijawab dengan

benar. Maka x adalah discrete random variable. Nilai yang

mungkin untuk x adalah :

x = {0, 1, 2, 3, 4, 5, 6}

Asuransi Prudential melakukan survey pada konsumennya untuk

menentukan jumlah anak berusia di bawah 22 tahun yang tinggal

di rumah.

Maka random variable adalah x yaitu jumlah anak yang tinggal di

rumah.

x = {0, 1, 2, 3, 4, ...} discrete random variable


Continuous Random Variable

A continuous random variable is a variable that can assume

any value on a continuum (can assume an uncountable

number of values)

thickness of an item

time required to complete a task

temperature of a solution

height, in inches

These can potentially take on any value, depending only on

the ability to measure accurately.


Discrete Random Variables

Can only assume a countable number of values

Examples:

Roll a die twice

Let x be the number of times 4 comes up

x = {0, 1, or 2 times}

Toss a coin 5 times

Let x be the number of heads

x = {0, 1, 2, 3, 4, or 5}


Discrete Probability Distribution

x Value Probability

0 1/4 = 0.25

1 2/4 = 0.50

2 1/4 = 0.25

5-9

Experiment: Toss 2 Coins. Let x = # heads.

T

T

4 possible outcomes

T

T

H

H

H H

Probability Distribution

0 1 2 x

0.50

0.25 P

rob

ab

ilit

y


A list of all possible [ xi , P(xi) ] pairs

xi = Value of Random Variable (Outcome)

P(xi) = Probability Associated with Value

xi’s are mutually exclusive (no overlap)

xi’s are collectively exhaustive (nothing left out)

0 ≤ P(xi) ≤ 1 for each xi

▪ The probability of xi is between 0 and 1

S P(xi) = 1

▪ The sum of all probabilities in the sample space = 1




3 possible outcomes 21 possible outcomes

The number of possible outcomes increase, the distribution besome smoother dan the individual probability of any particular

value tends to be reduce

Discrete Random Variable Mean

Expected Value (or mean) of a discrete distribution

(Weighted Average) the weight are probability

assigned ti the values.

The average value when the experiment that generates

values for the random varianle is repeated over the long

run. = E(x) = xi P(xi)

Where :E(x) = Expected value of xx = values of the random variableP(x) = probability of the randim variable taking on the value x


Discrete Random Variable Mean

Example: Toss 2 coins, x = # of heads,

compute expected value of x:

E(x) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25) = 1.0


x P(x)

0 0.25

1 0.50

2 0.25

Discrete Random Variable Standard Deviation

Variace of a discrete random variable

Standard Deviation of a discrete distribution

where:E(x) = Expected value of the random variable (x)

x = Values of the random variable

P(x) = Probability of the random variable having the value of x


)P(xE(x)}{xσ i

2

ix

)P(XE(X)][Xσ i

2

i

2

Discrete Random Variable Standard Deviation

Example: Toss 2 coins, x = # heads,

compute standard deviation (recall E(x) = 1)


)P(xE(x)}{xσ i

2

ix

.7070.50(0.25)1)(2(0.50)1)(1(0.25)1)(0σ 222

x

(continued)

Possible number of heads = 0, 1, or 2

Contoh

Karena cuaca buruk, jumlah hari pada minggu depan untuk

pelayaran menjadi tidak tentu. Misalkan x adalah hari

pelayaran terjadi. Berikut ini adalah distribusi probabilitas

dari variabel, x ditentukan berdasarkan data historis saat

cuaca buruk :

Berdasarkan distribusi probabilitas, tentukan expected dari

hari per minggu yang memungkinkan untuk pelayaran?

Tentukan pula standar deviasinya!


X 0 1 2 3 4 5 6 7

P(x) 0,05 0,1 0,1 0,2 0,2 0,15 0,15 0,05

Expected value :

Standar deviasi :


x 0 1 2 3 4 5 6 7

P(x) 0,05 0,1 0,1 0,2 0,2 0,15 0,15 0,05 Sum

x.P(x) 0 0,1 0,2 0,6 0,8 0,75 0,9 0,35 3,7

Probability Distributions

5-18

Binomial*

DiscreteProbability

Distributions

Normal

Uniform

Exponential

Ch. 5 Ch. 6

Binomial Negative

Hypergeometric*

Poisson*

ContinuousProbability

Distributions

Multinomial

Geometric

The Binomial Distribution

A distribution that gives the probability of x successes in n

trial in a process that meets the following condition

(characteristics the Binomial Distribution) :

A trial has only two possible outcomes – “success” or “failure”

There is a fixed number, n (finite), of identical trials

The trials of the experiment are independent of each other

The probability of a success, p, remains constant from trial to

trial

If p represents the probability of a success, then

(1-p) = q is the probability of a failure

The sampling is performed with replacement from a finite

population.


Binomial Distribution Examples

A manufacturing plant labels items as either defective or

acceptable

A firm bidding for a contract will either get the contract

or not

A marketing research firm receives survey responses of

“yes I will buy” or “no I will not”

New job applicants either accept the offer or reject it


Counting Rule for Combinations

A combination is an outcome of an experiment where x

objects are selected from a group of n objects.


)!xn(!x

!nCn

x

where:

Cx = number of combinations of x objects selected from n objects

n! =n(n - 1)(n - 2) . . . (2)(1)

x! = x(x - 1)(x - 2) . . . (2)(1)

n

NOTE: 0! = 1 (by definition)

Order does not matter

i.e. ABC = CBA only one outcome

Contoh

Dilakukan pemeriksaan terhadap 4 kelompok sepatu untuk

menentukan kualitas “good” dan “defect”.

Maka kombinasi cara untuk mendapatkan kemungkinan

“defect” bersifat discrete dengan hasil sebagai berikut :

x = {0, 1, 2, 3, 4}

x = 0

x = 1


1)1234(1

1234

)!04(!0

!44

0

xxx

xxxC

4)123(1

1234

)!14(!1

!44

1

xx

xxxC

G, G, G, G

G, G, G, D

G, G, D, G

G, D, G, G

D, G, G, G

x = 2

x = 3

x = 4


4)1(123

1234

)!34(!3

!44

3

xx

xxxC

1)1(1234

1234

)!44(!4

!44

4

xxx

xxxC

6)12(12

1234

)!24(!2

!44

2

xx

xxxC

G, G, D, D G, D, G, D

G, D, D, G D, G, D, G

D, D, G, G D, G, G, D

G, D, D, D

D, G, D, D

D, D, G, D

D, D, D, G

D, D, D, D

Binomial Distribution Formula


P(x) = probability of x successes in n trials, with probability of success p on each trial

x = number of successes in sample, (x = 0, 1, 2, ..., n)

p = probability of “success” per trialq = probability of “failure” = (1 – p)n = number of trials (sample size)

P(x)n

x ! n xp q

x n x!

( )!=

-

-

Example: Flip a coin four times, let x = #

heads:n = 4

p = 0.51 - p = (1 - 0.5) = 0.5

x = 0, 1, 2, 3, 4

Dengan menggunakan persoalan pada pemeriksaan sepatu

diketahui n = 4, p = 0,1 dan x = 2 defects

Maka :


0486,0)9,0)(1,0(!2!2

!4)2(

)!(!

!)(

22

P

qpxnx

nxP xnx

Distribution

x = # of defects

P(x)

0 0,6561

1 0,2916

2 0,0486

3 0,0036

4 0,0001

Total 1,0000


Perhitungan dengan menggunakan rumus seringkali memerlukan waktu.

Untuk itu dapat digunakan tabel Binomial.

Tabel Binomial


Example: Calculating a Binomial Probability

Business Statistics, A First Course (4e) © 2006 Prentice-Hall, Inc. 5-28

What is the probability of one success in five observations if the probability of success is .1?

x = 1, n = 5, and p = 0.1

0.32805

.9)(5)(0.1)(0

0.1)(1(0.1)1)!(51!

5!

p)(1px)!(nx!

n!1)P(x

4

151

nx

x

Binomial Distribution Summary Measures

Mean


Variance and Standard Deviation

npE(x)μ

npqσ2

npqσ

Where n = sample size

p = probability of success

q = (1 – p) = probability of failure

Binomial Distribution

The shape of the binomial distribution depends on the

values of p and n


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.2

.4

.6

0 1 2 3 4 5

X

P(X)

0

Here, n = 5 and p = 0.1

(Binomial distribution skewed. The skeweness will most pronoinced when n is small and p approaches 0 or 1)

Here, n = 5 and p = 0.5

(bell-shapes distribution)

Binomial Characteristics


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.2

.4

.6

0 1 2 3 4 5

X

P(X)

0

0.5(5)(0.1)npμ

0.6708

.1)(5)(0.1)(1npqσ

2.5(5)(0.5)npμ

1.118

.5)0(5)(0.5)(1npqσ

Examples

Binomial Distribution Example

Example: 35% of all voters support Proposition A. If a

random sample of 10 voters is polled, what is the probability

that exactly three of them support the proposition?

i.e., find P(x = 3) if n = 10 and p = 0.35 :


.25220(0.65)(0.35)3!7!

10!qp

x)!(nx!

n!3)P(x 73xnx

There is a 25.22% chance that exactly 3 out of the 10 voters will support Proposition A

n = 10

x p=.15 p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50

012345678910

0.19690.54430.82020.95000.99010.99860.99991.00001.00001.00001.0000

0.10740.37580.67780.87910.96720.99360.99910.99991.00001.00001.0000

0.05630.24400.52560.77590.92190.98030.99650.99961.00001.00001.0000

0.02820.14930.38280.64960.84970.95270.98940.99840.99991.00001.0000

0.01350.08600.26160.51380.75150.90510.97400.99520.99951.00001.0000

0.00600.04640.16730.38230.63310.83380.94520.98770.99830.99991.0000

0.00250.02330.09960.26600.50440.73840.89800.97260.99550.99971.0000

0.00100.01070.05470.17190.37700.62300.82810.94530.98930.99901.0000

Using Cumulative Binomial Distribution Tables

5-33Example: n = 10, p = 0.35, x = 3: P(x = 3, n =10, p = 0.35) = 0.5138

If a random sample of 10 voters is polled, what is the probability that three or fewer of them support the proposition?

Contoh

Manager PT Coca Cola Amatil mengirimkan 300 email kepada konsumen untuk mengukur kepuasan terhadap pelayanan tahunan. Selama ini tingkat pengembalian email (membalas email) adalah p = 0,11.

Sehingga expected value (mean) pengembalian email adalah : E(x) = n.p = 300 x 0,11 = 33

Pada tahun ini ternyata da 44 konsumen yang membalas email sebanyak 44 orang. Pengembalian ini melebihi 33 orang. Selanjutnya manager melakukan pengujian terhadap data tersebut : P(x≥44) = 1 – P(x≤43)

Tabel binomial memiliki keterbatasan untuk sample > 300, sehingga perhitungan dilakukan dengan menggunakan software excel ataupun minitab.


Using EXCEL

1. Buka sheet kosong

2. Pilih Formulas

3. Klik pada fx (Function

wizard)

4. Pilih Statistical Category

5. Pilih BINOMDIST

function

6. Lengkapi informasi yang

diminta

7. True mengindikasikan

probabilitas kumulatif


P(x≤43) = 0,97P(x≥44) = 1 – 0,97 = 0,03Hanya 3% peluang dari 44 atau lebih konsumen

yang membalas email dari peluang 11%.

The Poisson Distribution

Difficult to count the number of failures, so diffivult to

know possible outcomes (successes + failures).

Characteristics of the Poisson Distribution:

The outcomes of interest are rare relative to the possible

outcomes

The average number of outcomes of interest per time or

space interval is

The number of outcomes of interest are random, and the

occurrence of one outcome does not influence the chances

of another outcome of interest

The probability that an outcome of interest occurs in a

given segment is the same for all segmentsCopyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-36

Contoh :

Mendeskripsikan suatu situasi :

Jumlah pasien yang datang pada unit IGD per jam.

Jumlah panggilan pada customer service Telkom setiap

30 menit.

Mendeskripsikan random variable :

Jumlah ‘dot’ pada setuap lembar kertas.

Jumlah kontaminan dalam segalon air


Poisson Distribution Summary Measures

Mean

λ and t must be in compatibe unit.the average number in t segment, is not necessarily the number we eill see if we observe the process for t segments.


Variance and Standard Deviation

λtμ E(x) x

λtσ2

λtσ

where = number of successes in a segment of unit size

t = the size of the segment of interest

Graph of Poisson Probabilities


0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)X

t =

0.50

0

1

2

3

4

5

6

7

0.6065

0.3033

0.0758

0.0126

0.0016

0.0002

0.0000

0.0000P(x = 2) = 0.0758

Graphically:

= .05 and t = 100

Poisson Distribution Shape

The shape of the Poisson Distribution depends on the

parameters and t:


0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x

)

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)

t = 0.50 t = 3.0

Poisson Distribution Formula


where:

t = size of the segment of interest

x = number of successes in segment of interest

= expected number of successes in a segment of unit size

e = base of the natural logarithm system (2.71828...)

!

)()(

x

etxP

tx

Using Poisson Tables


Example: Find P(x = 2) if = 0.05 and t = 100

.075802!

e(0.50)

!

)()2(

0.502

x

etxP

tx

X

t

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

01234567

0.90480.09050.00450.00020.00000.00000.00000.0000

0.81870.16370.01640.00110.00010.00000.00000.0000

0.74080.22220.03330.00330.00030.00000.00000.0000

0.67030.26810.05360.00720.00070.00010.00000.0000

0.60650.30330.07580.01260.00160.00020.00000.0000

0.54880.32930.09880.01980.00300.00040.00000.0000

0.49660.34760.12170.02840.00500.00070.00010.0000

0.44930.35950.14380.03830.00770.00120.00020.0000

0.40660.36590.16470.04940.01110.00200.00030.0000

Contoh

Figur kedatangn dari konsumen Mc. D disajikan dlam

distribusi Poissson dengan λ = 16 konsume per jam.

Berdasarkan data didapatkan waktu pelayanan cukup

konstan pada 6 menit. Di restoran terdapat 3 kasir, sehingga

dalam 6 menit 3 konsumen dapat terlayani. Manager

bermaksud mengetahui probabilitas 1 atau 2 konsumen

tidak terlayani dalam 6 menit.

Convert 6 menit menjadi 0,1 jam sehingga t = 0,1. sehingga

λt = 16 (0,1) = 1,6 konsumen.

P(1 atau lebih konsumen menunggu) = P(4) + P(5) + P(6) + ...

Atau P(1 atau lebih konsumen menunggu) = 1 – P(x≤3)


P(x≤3) = 0,9212

Sehingga P(4 atau lebih konsumen) = 1 – 0,9212 = 0,0788

Manager bermaksud memenuhi target agar konsumen yang

menunggu tidak lebih dari 0,05 sebagai acuan untuk

penambahan kasir.

P(4 atau lebih konsumen) = 1 – P(x≤?) ≤0,05Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-44

Sehingga diperlukan P(x≤?) ≥0,95

P(4 atau lebih konsumen) = 1 – P(x≤4) = 1 – 0,9763 = 0,0237

Karena 0,0237 < 0,05 maka sebaiknya manager menambah

satu kasir.


The Hypergeometric Distribution

“n” trials in a sample taken from a finite population of

size N

Sample taken without replacement

Trials are dependent

The probability changes from trial to trial

Concerned with finding the probability of “x” successes

in the sample where there are “X” successes in the

population


Hypergeometric Distribution Formula


N

n

X

x

XN

xn

C

CCxP

)(.

Where

N = population size

X = number of successes in the population

n = sample size

x = number of successes in the sample

n – x = number of failures in the sample

(Two possible outcomes per trial: success or failure)

Hypergeometric Distribution Example

■ Example: 3 Light bulbs were selected from 10. Of the 10

there were 4 defective. What is the probability that 2 of

the 3 selected are defective?

N = 10 n = 3

X = 4 x = 2


0.3120

(6)(6)

C

CC

C

CC2)P(x

10

3

4

2

6

1

N

n

X

x

XN

xn

Contoh

Pada produksi Nokia bulan Juli didapatkan 20 produk dicurigai

defect karena kesalahan proses. Manager segera memerintahkan

untuk memisahkan 20 produk tersebut. Namun bagian QC

terlanjur meloloskan 10 produk ke bagian pengepakan. Peluang

defect dari 20 product adalah 2.

Maka peluang produk tanpa defect yang dikirim adalah (x = 0)?


0.2368184,756

)(43,758)(10)P(x

C

CC

C

CC0)P(x

20

10

2

0

2-20

0-10

N

n

X

x

XN

xn

Hypergeometric dustribution untuk semua defect yang

mungkin :


x P(x)

0 0,2368

1 0,5264

2 0,2368

Total 1,0000

Hypergeometric Distribution (k possible outcomes per trial)

N = Population size

n = Total sample size

Xi = Number of items in the

population with outcome i

xi = Number of items in the sample

with outcome i


N

n

X

x

X

x

X

x

kC

CCCxxxP

K

k...

),...,,(2

2

1

1

21

nx

NX

k

i

i

k

i

i

1

1

Contoh

Studi terhadapt pasar sepeda motor dengan 4 merek yang ada pada suatu dealer. Pada dealer tersebut

tersebut terdapat stok sebagai berikut : 5 brand Honda

4 brand Suzuki

6 brand Yamaha

4 brand Kawasaki

Sales manager melakukan penelitian pilihan konsumen tanpa melihat harga. Sebanyak 6 konsumen dipilih secara acak dan 3 memilih Suzuki, 2 memilih Kawasaki dan 1 memilih Yamaha. Tidak satupun memilih Honda.

Propabilitas pemilihan secara acak dengan asumsi

memilih tanpa pengembalian adalah :


Tentukan ukuran populasi dan sample :

N = 19 dan n = 6

Tentukan event of interest :

P(x1 = 0, x2 = 3, x3 = 1, x4 =2) = .....

Tentukan jumlah tiap kategori tiap populasi dan sample

:


Honda Suzuki Yamaha Kawasaki Total

X1 = 5 X2= 4 X3 = 6 X4 = 4 N = 19

x1 = 0 X2 = 3 x3 = 1 x4 = 2 n = 6

Hitung probabilitas


Chapter Summary

Reviewed key discrete distributions

Binomial

Poisson

Hypergeometric

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems


Documents

# 6 Distribusi Probabilitas Diskrit